Question

Calculate the pH of the cathode compartment for thefollowing reaction given cell = 3.01 V when [Cr³⁺] =0.15 M, [A¹³⁺] = 0.30 M, and [Cr₂O₇^2-] = 0.55 M.

$\mathbf{2}\mathit{A}\mathit{l}\left(s\right)\mathbf{+}\mathit{C}{\mathit{r}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{7}}}^{\mathbf{2}\mathbf{-}}\left(aq\right)\mathbf{+}\mathbf{14}{\mathit{H}}^{\mathbf{+}}\left(aq\right)\mathbf{\to }\mathbf{2}\mathit{A}{\mathit{l}}^{\mathbf{3}\mathbf{+}}\left(aq\right)\mathbf{+}\mathbf{2}\mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}\left(aq\right)\mathbf{+}\mathbf{7}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(l\right)$

Short answer

The value of pH of the cathode compartment is0.05.

Step-by-step solution

Half-cell reaction at anode and cathode

The Half-cell reaction at anode and cathode can be denoted as:

$$\begin{gathered} 2Al\left(s\right)\to 2A{l}^{+3}\left(aq\right)+6e^{-} \\ C{r}_2{O_7}^{2-}\left(aq\right)+14H^{+}\left(aq\right)+6e^{-}\to 2C{r}^{+3}\left(aq\right)+7H_{2}O \end{gathered}$$

The overall cell potential can be calculated using,

${E^o}_{cell}={E^o}_{cathode}-{E^o}_{anode}$

On substituting the values,

$$\begin{gathered} {E^o}_{cell}=1.33 V-(-1.66 V) \\ =2.99 V \end{gathered}$$

Expression of reaction quotient

The expression for reaction quotient can be expressed as:

$$\begin{gathered} Q=\frac{{\left[A{l}^{+3}\right]}^2{\left[C{r}^{+3}\right]}^2}{\left[C{r}_2{O_7}^{2-}\right]{\left[H^{+}\right]}^{14}} \\ Q=\frac{{(0.6 M)}^2{(0.3 M)}^2}{(0.55 M){\left[H^{+}\right]}^{14}} \\ Q=\frac{0.059}{{\left[H^{+}\right]}^{14}} \end{gathered}$$

Calculate the value of pH

The formula can be expressed as:

$$\begin{gathered} E_{cell}={E^o}_{cell}-\frac{0.0591}{n}\log \left(Q\right) \\ 3.01-2.99=-0.0099\log \left(\frac{0.059}{{\left[H^{+}\right]}^{14}}\right) \\ -2.02=-1.22-14pH \\ pH=0.05 \end{gathered}$$

The value of pH of the cathode compartment is 0.05.